Deriving the 206265 instead of quoting it
A pixel of width p sitting at a distance F behind a lens subtends an angle of p divided by F, in radians, for any angle small enough that the tangent and the angle agree — which every pixel on every telescope comfortably is. That is the entire physics. The rest is unit conversion.
One radian is 180 divided by pi degrees, which is 57.29578 degrees. Multiply by 3600 to get arcseconds and you have 206264.806 arcseconds in a radian. So the angle in arcseconds is 206264.806 times p divided by F. Put p in microns and F in millimetres and the factor of a thousand between the units cancels three of the digits, leaving the familiar form: scale equals 206.265 times pixel size in microns, divided by focal length in millimetres.
Worth doing once by hand, because the constant then stops being a thing to look up and starts being a thing you can reconstruct. A 3.76 micron pixel at 1000 mm: 206.265 times 3.76 is 775.6, divided by 1000 gives 0.776 arcseconds per pixel.
Field of view is a different question with the same inputs
Multiply the scale by the pixel count and you have the field. A row of 6248 pixels at 0.776 arcseconds each is 4846 arcseconds, which is 80.8 arcminutes, which is 1.35 degrees. That is a wide field by deep sky standards and it will hold most large nebulae; it will not hold the Andromeda galaxy, which runs past three degrees.
Note that binning does not change the field. Binning combines adjacent pixels, so the scale per stored pixel goes up by the bin factor while the physical chip still covers the same sky. The frame gets smaller in pixels, not in degrees. This is the point people most often trip over when comparing two cameras, and it is why the calculator computes the field from the native pixel size and reports the binned pixel count separately.
Sampling, seeing, and why there is no correct answer
A star does not arrive as a point. The atmosphere smears it into a blob, conventionally measured as the full width at half maximum of that blob, and the width changes with the night, the altitude of the target and the hour. On a good site on a good night it might be well under two arcseconds; from a suburban garden on a mediocre night it can be four or more. That is why the seeing figure on this page is something you enter from your own measurement rather than something the page asserts.
| Samples across the seeing disc | What you gain | What you pay |
|---|---|---|
| Around 1 | Deep frames fast, small files, forgiving of tracking error | Stars go blocky, fine structure is gone for good |
| Around 2 | Most of the detail the night offered | Little, on a night that never gets steady |
| Around 3 | Room for deconvolution and drizzle to work | Each pixel gets a third of the light, so integration has to grow |
| Above 4 | Nothing, on most nights from most sites | Noise, storage, and exposure time spent on empty resolution |
The argument over where in that range to sit is long-running and both sides have a case. What is not in dispute is the arithmetic: halving the scale quarters the light per pixel, so the same signal-to-noise costs four times the integration. That is the real price of oversampling and it is why the calculator reports the focal length that would hit your chosen target rather than declaring a target for you.
Binning is not what it used to be
On a CCD, binning happened in the readout register before the noise was added, so combining four pixels genuinely gave four times the signal for one dose of read noise. On most modern CMOS sensors the pixels are read individually and binning is arithmetic done afterwards, which still helps — averaging four values reduces the noise — but it does not help in the same way or by the same amount. The scale and field arithmetic on this page is identical either way. The signal-to-noise argument is not, and the sensor design is what decides it.
The practical version: binning after the fact costs you nothing, because you can always do it later to a stored frame. Binning in hardware at capture time is irreversible. When storage rather than noise is the constraint, the file size calculator and the integration planner are the pages that turn frame counts into gigabytes.
Diffraction sets the floor the atmosphere sits on
Even with no atmosphere the aperture limits resolution, because light passing through a circular opening spreads into a disc with rings around it. The Rayleigh criterion puts two stars at the limit of separation when the centre of one falls on the first dark ring of the other, which works out at 1.22 times the wavelength divided by the aperture, in radians. At 550 nanometres through 200 mm that is 0.69 arcseconds.
The Dawes limit is a different kind of number. It is an empirical fit, made by William Dawes from what observers could actually split on pairs of equal-brightness stars, and it comes out as 116 divided by the aperture in millimetres — tighter than Rayleigh, because the eye can detect an elongation before it can see two separate discs. This page reports Rayleigh as the derived physical limit and Dawes beside it as the observational convention, and it names which is which, because quoting one under the other name is the standard way this gets muddled.
For deep sky imaging both are usually academic: at two arcseconds of seeing, a 0.69 arcsecond diffraction limit is buried three times over. On planetary work at short exposures where the atmosphere can be frozen, it is the number that matters.
Questions people ask
How do I calculate arcseconds per pixel?
Multiply the pixel size in microns by 206.265 and divide by the focal length in millimetres. The constant is not arbitrary — it is the number of arcseconds in a radian, 180 times 3600 divided by pi, which is 206264.806, with a factor of a thousand absorbed by mixing microns and millimetres. A 3.76 micron pixel at 1000 mm gives 0.776 arcseconds per pixel. Reconstruct the constant once and you will not need to look it up again.
What sampling should I aim for?
There is no figure that is right for everyone, which is why this page takes the target as an input. The argument runs roughly between two and three-something samples across the seeing disc, and where you land depends on your sky, your mount and whether you intend to deconvolve. What is not arguable is the cost: halving the arcseconds per pixel quarters the photons landing in each one, so the same signal-to-noise needs four times the total integration. Enter your own measured seeing and your own target and read off the focal length that matches.
Does binning change my field of view?
No. Binning combines neighbouring pixels, so each stored pixel covers more sky and the frame contains fewer pixels, but the physical sensor still sits behind the same optics covering the same angle. Two by two binning doubles the arcseconds per pixel and quarters the pixel count, leaving the field exactly where it was. This page computes the field from the native pixel size for that reason and reports the binned dimensions separately.
What is the difference between the Dawes limit and the Rayleigh criterion?
Rayleigh is derived from diffraction: two point sources are at the limit when the centre of one Airy disc falls on the first dark ring of the other, giving 1.22 times wavelength over aperture in radians. Dawes is empirical, fitted in the 1860s to what observers could actually split on equal-brightness double stars, and works out as 116 divided by the aperture in millimetres. Dawes is the tighter number because a trained eye detects elongation before it sees two discs. Both appear on this page, labelled, because double star catalogues are written against Dawes while the physics is Rayleigh.
My stars look bloated. Is that a sampling problem?
Sampling sets the smallest star you could record; it does not create big ones. Bloated stars usually come from something else — focus drift as the tube cools, tracking or guiding error smearing each sub, atmospheric seeing that was worse than the night looked, or aggressive stretching in processing pulling faint halos up into view. Work out your scale here, then compare the measured star width in your frames against the seeing you entered. If the frames are much worse than the seeing, the problem is mechanical or optical rather than a sampling choice.